Can you please solve Question 4? 1) The tires on a new car have a diameter of 2 ft and have a 60,000-mile warranty.a) Determine the angle (in radians) through which one of these tires will rotate during thewarranty period.b) How many revolutions is this?2) A centrifuge in a medical laboratory rotates at an angular speed of 3200 rev/min. Whenswitched off, it rotates through 60 revolutions before coming to rest. Find the constant angularacceleration of the centrifuge.3) Find the angular speed of Earth about the Sun in radians per second and degrees per day.4) Find the centripetal accelerations ofa) a point on the equator of Earthb) the North Pole, andc) New York,due to the rotation of Earth about its axis.

Given data *The radius of the earth is r = 6.38 × 106 m *The period is T = 1 day = 86400 s *The…

Answered: 4) Find the centripetal accelerations…

College Physics

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ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

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Can you please solve Question 4?

1) The tires on a new car have a diameter of 2 ft and have a 60,000-mile warranty.
a) Determine the angle (in radians) through which one of these tires will rotate during the
warranty period.
b) How many revolutions is this?
2) A centrifuge in a medical laboratory rotates at an angular speed of 3200 rev/min. When
switched off, it rotates through 60 revolutions before coming to rest. Find the constant angular
acceleration of the centrifuge.
3) Find the angular speed of Earth about the Sun in radians per second and degrees per day.
4) Find the centripetal accelerations of
a) a point on the equator of Earth
b) the North Pole, and
c) New York,
due to the rotation of Earth about its axis.

Expert Solution

Step 1

Given data

*The radius of the earth is r = 6.38 × 10⁶ m

*The period is T = 1 day = 86400 s

*The radius of the earth at the north pole is R = 6.35 × 10⁶ m

Step 2

(a)

The expression for the centripetal acceleration at a point on the equator is given as

$\begin{array}{rcl} a_{c} & = & ω^{2} r \\ = & {(2 πf)}^{2} r \\ = & {(\frac{2 π}{T})}^{2} r \end{array}$

Substitute the values in the above expression as

$\begin{array}{rcl} a_{c} & = & {(\frac{2 \times 3.14}{86400 s})}^{2} \times (6.38 \times 10^{6} m) \\ = & 3.37 \times 10^{- 2} m / s^{2} \end{array}$

Thus the centripetal acceleration at a point on the equator is a_c = 3.37 × 10^-2 m/s²

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